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Question
If tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\]
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Solution
\[\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\]
\[ = \frac{\frac{1}{1 + 2^{- x}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{(1 + 2^{- x} )(1 + 2^{x + 1} )}}\]
\[ = \frac{1 + 2^{x + 1} + 1 + 2^{- x}}{1 + 2^{x + 1} + 2^{- x} + 2^{- x + x + 1} - 1}\]
\[ = \frac{2 + 2^{x + 1} + 2^{- x}}{2 + 2^{x + 1} + 2^{- x}}\]
\[ = 1\]
\[\text{ Therefore }, \alpha + \beta = \tan^{- 1} (1) = \frac{\pi}{4} .\]
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